Gujarat Board GSEB Solutions Class 10 Maths Chapter 8 Introduction to Trigonometry Ex 8.3 Textbook Questions and Answers.

Gujarat Board Textbook Solutions Class 10 Maths Chapter 8 Introduction to Trigonometry Ex 8.3

Question 1.
(i) (frac{sin 18^{circ}}{cos 72^{circ}})
(ii) (frac{tan 26^{circ}}{cot 64^{circ}})
(iii) cos 48° – sin 42°
(iv) cosec 31° – sec 59°
Solution:
(i) (frac{sin 18^{circ}}{cos 72^{circ}}) = (frac{sin 18^{circ}}{cos left(90^{circ}-18^{circ}right)})
= sin (frac{sin 18^{circ}}{sin 18^{circ}}) = 1 (cos (90° – θ) = sin θ)

(ii) (frac{sin 18^{circ}}{cos 72^{circ}}) = (frac{tan 26^{circ}}{cot left(90^{circ}-26^{circ}right)})
= (frac{tan 26^{circ}}{tan 26^{circ}}) = 1 (cot(90°- θ) = tan θ)

(iii) cos 48° – sin 42°
= cos (90° – 42°) – sin 42°
= sin 42° – sin 42° = 0 [cos (90° – θ) = sin θ]

(iv) cosec 31° – sec 59°
= cos (90° – 59°) – sec 59°
= sec 59° – sec 59° = 0

Question 2.
Show that
(i) tan 48° tan 23° tan 42° tan 67° = 1
(ii) cos 38° cos 52° – sin 38° sin 52° = 0
Solution:
(i) LHS = tan 48° tan 23° tan 42° tan 67°
= tan 48° tan 42° tan 23° tan 67°
= tan (90° – 42°) tan 42° tan 23° tan(90° – 23°)
= cot 42° tan 42° tan 23° cot 23°
= (frac {1}{tan 42°}) x tan 42° x tan 23° x (frac {1}{tan 23°})
=1 = RHS

(ii) LHS = cos 38° cos 52° – sin 38° sin 52°
= cos 38° a (90° – 38°) – 38° sin (90° – 38°)
= cos 38° sin 38° – sin 38° cos 38° = 0 = RHS

Question 3.
If tan 2A = cot (A – 18°), where 2A is an acute angle, find the value of A. (CBSE 2012)
Solution:
We have
tan 2A cot (A – 18°)
tan 2A = tan [90°- (A – 18°)]
tan 2A = tan [90° – A + 18°]
2A = 90° – A + 18°
3A = 108°
A = (frac {108°}{3})
A = 36°

Question 4.
If tan A = cot B, prove that A+ B = 90°.
Solution:
We have
tan A = cot B
tanA = tan(90° – B) [cot θ = tan (90° – θ)]
A = 90°- B
A + B = 90°

Question 5.
If sec 4A = cosec (A – 20°) where 4A is an acute angle, find the value of A.
Solution:
We have
sec 4A = cosec (A – 20°)
cosec (90° – 4A) = cosec (A – 20°)
90° – 4A = A – 20°
90° + 20° = 5A
5A = 110°
A = (frac {110°}{5}) = 22°

Question 6.
If A, B and C are interior angles of a triangle ABC, then show that
sin (frac {B + C}{2}) = cos = (frac {A}{2})
Solution:
In a triangle,
A + B + C = 180°
(Sum of angles of a triangle is 180°)
(frac {A}{2}) + (frac {B}{2}) + (frac {C}{2}) = (frac {180°}{2})
(Dividing by 2 on both sides)
⇒ (frac {B}{2}) + (frac {C}{2}) = 90° – (frac {A}{2})
(frac {B + C}{2}) = 90° – (frac {A}{2})
sin ((frac {B + C}{2})) = sin ((left(90^{circ}-frac{mathrm{A}}{2}right)))
(Taking sin on both sides)
sin ((frac {B + C}{2})) = cos (frac {A}{2})

Question 7.
Express sin 67° ÷ cos 75° in terms of trigonometric ratios of angles between 0° and 45°.
Solution:
We have
sin 67° + cos 75°
= sin (90° – 23°) + cos (90° – 15°)
= cos 23° + sin 15°